You are growing yeast for a science experiment. You start
with 10 ml. You set the experiment up, go away and return 45
minutes later. When you come back, your have 30 ml. Find the
rate at which it is growing.
The EXACT answer (no calculator necessary) can be given
in the form a ln(b).
So I do know the answer to this, I got it wrong on the test i took last week, but I am hoping someone can explain to me the process of how to get the answer.
I had used to A=(1+r/n)^(nt) but i ended up getting so lost in the problem that I left it blank

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You are growing yeast for a science experiment. You start
with 10 ml. You set the experiment up, go away and return 45
minutes later. When you come back, your have 30 ml. Find the
rate at which it is growing.
The EXACT answer (no calculator necessary) can be given
in the form a ln(b).
So I do know the answer to this, I got it wrong on the test i took last week, but I am hoping someone can explain to me the process of how to get the answer.
I had used to A=(1+r/n)^(nt) but i ended up getting so lost in the problem that I left it blank

Mathematics

Stacey Warren - Expert brainly.com

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More answers

so here is what you know
when you start you have 10, so you begin with \(10e^{rt}\)

anonymous

5 years ago

you also know that when \(t=45\) you have 30, so replace A by 30 and \(t\) by 45 and solve for \(r\)

anonymous

5 years ago

i.e. solve
\[30=10e^{45r}\] for \(r\)

anonymous

5 years ago

you good from there?
divide by 10 get
\[3=e^{45r}\] then take the log and finally divide by 45

anonymous

5 years ago

you could have actually started at that step, reasoning that it triple is 45 minutes

anonymous

5 years ago

so i would have log3= loge^45r ?

anonymous

5 years ago

thank you so much for explaining this to me! :)

anonymous

5 years ago

no you would have
\[\log(3)=45r\]

anonymous

5 years ago

i though what you did to once side you did to the other?

anonymous

5 years ago

when you take the log of \(e^{45r}\) you just get \(45r\)

anonymous

5 years ago

oh that's right. there are so many rules

anonymous

5 years ago

\[e^x=y\iff x=\ln(y)\] they say the same thing

anonymous

5 years ago

well i wouldn't call that a "rule" it is equivalent logarithmic form
or how one gets a variable out of the exponent

anonymous

5 years ago

don't forget you are trying to solve for a variable that is in the sky (exponent) algebra will not do that for you
you need logs for that

anonymous

5 years ago

to me they all feel like rules and properties right now, probably until get the hang of it more

anonymous

5 years ago

that makes sense. thank you so much! i hate when i get a problem wrong on a test and i can't figure out where i went wrong.

anonymous

5 years ago

hope it is clear though. you are looking for \(r\) and it is an exponent
that is why this is different from algebra.
you need to take the log to get it

anonymous

5 years ago

you remember finding the equation of a line given two points? this is analogous to that
\(b\) is the \(y\) intercept, what you get when \(x=0\)
in this case \(P\) is the \(y\) intercept, what you get when \(t=0\)

anonymous

5 years ago

yes. i do.

anonymous

5 years ago

in a line you needed to find
\(m\)
in this case you need to find \(r\)

anonymous

5 years ago

that actaully makes so much sense to me. i am always trying to bring the r and t down like numerical exponents and solving them algebraically. .. which is usually wrong

anonymous

5 years ago

and you do it almost like you do with a line
if you have \(y=4x+b\) and you did not know \(b\) you could find it if say \((3,5)\) is on the graph by saying
\[5=4\times 3+b\] and solving for \(b\)
this is almost the same
we know \((45,30)\) is on the graph so you can solve
\[30=10e^{45r}\] for \(r\) using the log